AIME 2001 I · 第 11 题

Solution

Let each point $P_i$ be in column $c_i$. The numberings for $P_i$ can now be defined as follows.

$$\begin{aligned}x_i &= (i - 1)N + c_i\\ y_i &= (c_i - 1)5 + i \end{aligned}$$ We can now convert the five given equalities.

$$\begin{aligned}x_1&=y_2 & \Longrightarrow & & c_1 &= 5 c_2-3\\ x_2&=y_1 & \Longrightarrow & & N+c_2 &= 5 c_1-4\\ x_3&=y_4 & \Longrightarrow & & 2 N+c_3 &= 5 c_4-1\\ x_4&=y_5 & \Longrightarrow & & 3 N+c_4 &= 5 c_5\\ x_5&=y_3 & \Longrightarrow & & 4 N+c_5 &= 5 c_3-2 \end{aligned}$$ Equations $(1)$ and $(2)$ combine to form

$$N = 24c_2 - 19$$ Similarly equations $(3)$, $(4)$, and $(5)$ combine to form

$$117N +51 = 124c_3$$ Take this equation modulo 31

$$24N+20\equiv 0 \pmod{31}$$ And substitute for N

$$24 \cdot 24 c_2 - 24 \cdot 19 +20\equiv 0 \pmod{31}$$ $$18 c_2 \equiv 2 \pmod{31}$$ Thus the smallest $c_2$ might be is $7$ and by substitution $N = 24 \cdot 7 - 19 = 149$

The column values can also easily be found by substitution

$$\begin{aligned}c_1&=32\\ c_2&=7\\ c_3&=141\\ c_4&=88\\ c_5&=107 \end{aligned}$$ As these are all positive and less than $N$, $\boxed{149}$ is the solution.

Sidenote

If we express all the $c_i$ in terms of $N$, we have

$$24c_1=5N+23$$ $$24c_2=N+19$$ $$124c_3=117N+51$$ $$124c_4=73N+35$$ $$124c_5=89N+7$$ It turns out that there exists such an array satisfying the problem conditions if and only if

$$N\equiv 149 \pmod{744}$$ In addition, the first two equation can be written $n = 5mod24$, and chasing variables in the last three equation gives us $89n + 7 = 124e$. With these two equations you may skip a lot of rewriting and testing. $\boxed{149}$ still appears as our answer.

-jackshi2006